$h(t) = 3t^{2}-4t+3(f(t))$ $g(x) = -x^{2}+4x$ $f(n) = -5n^{3}+2n^{2}+2n-3-3(g(n))$ $ g(f(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = -5(1^{3})+2(1^{2})+(2)(1)-3-3(g(1))$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = -1^{2}+(4)(1)$ $g(1) = 3$ That means $f(1) = -5(1^{3})+2(1^{2})+(2)(1)-3+(-3)(3)$ $f(1) = -13$ Now we know that $f(1) = -13$ . Let's solve for $g(f(1))$ , which is $g(-13)$ $g(-13) = -(-13)^{2}+(4)(-13)$ $g(-13) = -221$